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<h1 class="heading"><a href="MATH-2023-OPDE.html"><span class="title">MATH 2023: Ordinary and Partial Differential Equations</span></a></h1>
<p class="byline">Xiaoyi Chen and Wei Zhang</p>
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<a href="ch_first.html" data-scroll="ch_first" class="internal"><span class="codenumber">1</span> <span class="title">Introduction</span></a><ul>
<li><a href="sec_1-intro.html" data-scroll="sec_1-intro" class="internal">Classification of Differential Equations</a></li>
<li><a href="sec_2-intro.html" data-scroll="sec_2-intro" class="internal">Linear and Nonlinear Equation</a></li>
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<a href="ch_second.html" data-scroll="ch_second" class="internal"><span class="codenumber">2</span> <span class="title">First Order Ordinary Differential Equations</span></a><ul>
<li><a href="sec2_1.html" data-scroll="sec2_1" class="internal">Linear Equations</a></li>
<li><a href="sec2_2.html" data-scroll="sec2_2" class="internal">Further Discussion of Linear Equations (For reading only)</a></li>
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<a href="ch_third.html" data-scroll="ch_third" class="internal"><span class="codenumber">3</span> <span class="title">third Order Linear Equations</span></a><ul>
<li><a href="sec3_1.html" data-scroll="sec3_1" class="internal">Homogeneous equations with constant coefficient</a></li>
<li><a href="sec3_2.html" data-scroll="sec3_2" class="internal">Fundamental Solutions of Linear Homogeneous Equations</a></li>
<li><a href="sec3_3.html" data-scroll="sec3_3" class="internal">Linear Independence and Wronskian</a></li>
<li><a href="sec3_4.html" data-scroll="sec3_4" class="internal">Complex roots of the characteristic equations</a></li>
<li><a href="sec3_5.html" data-scroll="sec3_5" class="internal">Repeated Roots: Reduction of Order</a></li>
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<li><a href="sec4_1.html" data-scroll="sec4_1" class="internal">General Theory of the <span class="process-math">\(n\)</span>-th Order Linear Equations</a></li>
<li><a href="sec4_2.html" data-scroll="sec4_2" class="internal">Homogeneous Equations with Constant Coefficients</a></li>
<li><a href="sec4_3.html" data-scroll="sec4_3" class="internal">The Method of Undetermined Coefficients</a></li>
<li><a href="sec4_4.html" data-scroll="sec4_4" class="internal">The Method of Variation of Parameters</a></li>
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<a href="ch_five.html" data-scroll="ch_five" class="internal"><span class="codenumber">5</span> <span class="title">Series Solutions of Second Order Linear Equations</span></a><ul>
<li><a href="sec5_1.html" data-scroll="sec5_1" class="internal">Brief Review on Power Series</a></li>
<li><a href="sec5_2.html" data-scroll="sec5_2" class="internal">Introduction</a></li>
<li><a href="sec5_3.html" data-scroll="sec5_3" class="internal">Series Solutions Near an Ordinary Point</a></li>
<li><a href="sec5_4.html" data-scroll="sec5_4" class="internal">Euler’s Equation</a></li>
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<a href="ch_six.html" data-scroll="ch_six" class="internal"><span class="codenumber">6</span> <span class="title">System of First Order Linear Equations</span></a><ul>
<li><a href="sec6_1.html" data-scroll="sec6_1" class="internal">Introduction <span class="process-math">\(\&amp;\)</span> Basic Theory</a></li>
<li><a href="sec6_2.html" data-scroll="sec6_2" class="internal">Homogeneous System with Constant Coefficients</a></li>
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<a href="ch_seven.html" data-scroll="ch_seven" class="internal"><span class="codenumber">7</span> <span class="title">Partial Differential Equations</span></a><ul>
<li><a href="sec7_1.html" data-scroll="sec7_1" class="internal">Two-Point Boundary Value Problems</a></li>
<li><a href="sec7_2.html" data-scroll="sec7_2" class="internal">Eigenvalue Problems</a></li>
<li><a href="sec7_3.html" data-scroll="sec7_3" class="internal">Fourier Series</a></li>
<li><a href="sec7_4.html" data-scroll="sec7_4" class="internal">The Fourier Convergence Theorem</a></li>
<li><a href="sec7_5.html" data-scroll="sec7_5" class="internal">Even and Odd Functions</a></li>
<li><a href="sec7_6.html" data-scroll="sec7_6" class="internal">Introduction to Partial Differential Equations</a></li>
<li><a href="sec7_7.html" data-scroll="sec7_7" class="internal">1D Heat Equation; Solutions by Separation of Variable and Fourier Series</a></li>
<li><a href="sec7_8.html" data-scroll="sec7_8" class="internal">Other Heat Conduction Problems</a></li>
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<a href="ch_eight.html" data-scroll="ch_eight" class="internal"><span class="codenumber">8</span> <span class="title">Laplace transform</span></a><ul>
<li><a href="sec8_1.html" data-scroll="sec8_1" class="internal">What are Laplace Transforms, and Why?</a></li>
<li><a href="sec8_2.html" data-scroll="sec8_2" class="internal">Finding Laplace Transforms</a></li>
<li><a href="sec8_3.html" data-scroll="sec8_3" class="internal">Finding inverse transforms using partial fractions</a></li>
<li><a href="sec8_4.html" data-scroll="sec8_4" class="internal">Solving ODEs and ODE Systems</a></li>
<li><a href="sec8_5.html" data-scroll="sec8_5" class="internal">Step input and Impulse problems</a></li>
<li><a href="sec8_6.html" data-scroll="sec8_6" class="internal">Laplace transform for PDE (heat equation)</a></li>
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<li class="link"><a href="solutions-1.html" data-scroll="solutions-1" class="internal"><span class="codenumber">A</span> <span class="title">Selected Hints</span></a></li>
<li class="link"><a href="solutions-2.html" data-scroll="solutions-2" class="internal"><span class="codenumber">B</span> <span class="title">Selected Solutions</span></a></li>
<li class="link"><a href="appendix-1.html" data-scroll="appendix-1" class="internal"><span class="codenumber">C</span> <span class="title">List of Symbols</span></a></li>
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<main class="main"><div id="content" class="pretext-content"><section class="section" id="sec2_5"><h2 class="heading hide-type">
<span class="type">Section</span> <span class="codenumber">2.5</span> <span class="title">Applications of modeling with first order ODE(For reading only)</span>
</h2>
<p id="p-36">First we review some results in mechanics.1. Newton’s second law:This law gives <span class="process-math">\(F=ma\text{,}\)</span> where <span class="process-math">\(F\)</span> is the external force, <span class="process-math">\(m\)</span> is the mass and <span class="process-math">\(a\)</span> is the acceleration. The basic units are <span class="process-math">\(s\)</span> for time, <span class="process-math">\(m\)</span> for length and <span class="process-math">\(kg\)</span> for mass. The unit for the force is <span class="process-math">\(N\)</span> and for the acceleration is <span class="process-math">\(m/s^2\text{.}\)</span>2. Gravity of earthGravity is given by the formula</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\textrm{Gravity}=-\frac{mg R^2}{(R+x)^2}=-\frac{mg}{\left(1+\frac{x}{R}\right)^2},
\end{equation*}
</div>
<p class="continuation">where <span class="process-math">\(g=9.8 m /s^2\text{.}\)</span> For small <span class="process-math">\(x\text{,}\)</span> we have Taylor expansion of <span class="process-math">\((1+\frac{x}{R})^{-2}\)</span> so that</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\textrm{Gravity}\approx -mg (1-2 \frac{x}{R}+\cdots)\approx -mg.
\end{equation*}
</div>
<p class="continuation">3. Other external force For example, the air resistance can be regarded as being proportional to the velocity, say <span class="process-math">\(-kv\)</span> where <span class="process-math">\(k\)</span> is a known constant.<dfn class="terminology">Example 1</dfn> A body of constant mass <span class="process-math">\(m\)</span> is projected vertically upward from the surface of the earth with an initial velocity <span class="process-math">\(v_0\text{.}\)</span> The gravitational acceleration of the earth is assumed to be constant. During the motion, the body is subjected to an air resistance which is proportional to the magnitude of the velocity, say, <span class="process-math">\(k |v|\text{.}\)</span> Finda) The time at which the maximum height is reached;b) The maximum height attained by the body.<dfn class="terminology">Solution:</dfn> From Newton’s second law, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&amp;F=-kv-mg=ma=m \frac{\textrm{d} v}{\textrm{d} t},\\
&amp; v(0)=v_0.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">We want to solve the above initial value problem. The differential equation is separable equation and it is rewritten as</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
-\textrm{d} t=\frac{\textrm{d} v}{\frac{k}{m} v+g}.
\end{equation*}
</div>
<p class="continuation">Integrating on both sides gives</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
-t+C_1=\frac{m}{k} \ln |\frac{k}{m}v+g|~\rightarrow~|\frac{k}{m}v+g|=e^{\frac{k}{m} C_1} e^{-\frac{k}{m} t}~\rightarrow~\frac{k}{m} v+g=\pm e^{\frac{k}{m} C_1} e^{-\frac{k}{m} t}=C e^{-\frac{k}{m} t}.
\end{equation*}
</div>
<p class="continuation">So the general solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
v=\frac{m}{k}\left(C e^{-\frac{k}{m} t}-g \right).
\end{equation*}
</div>
<p class="continuation">Using the initial condition, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="" id="eq2_18">
\begin{equation}
v_0=\frac{m}{k} (C-g)~\rightarrow~C=\frac{k}{m} v_0+g.\tag{2.5.1}
\end{equation}
</div>
<p class="continuation">Therefore, the solution to the initial value problem is</p>
<div class="displaymath process-math" data-contains-math-knowls="" id="eq2_18_1">
\begin{equation}
v=\frac{m}{k}\left[\left(\frac{k}{m} v_0+g\right) e^{-\frac{k}{m} t}-g \right]=\left(v_0+\frac{mg}{k} \right) e^{-\frac{k}{m} t}-\frac{mg}{k}.\tag{2.5.2}
\end{equation}
</div>
<p id="p-37">Further, one has the system governing the displacement as</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_19.html" id="eq2_19">
\begin{equation}
\begin{aligned}
&amp;\frac{\textrm{d} x}{\textrm{d} t}=v=\left(v_0+\frac{mg}{k} \right) e^{-\frac{k}{m} t}-\frac{mg}{k},\\
&amp;x(0)=0.
\end{aligned}\tag{2.5.3}
\end{equation}
</div>
<p class="continuation">From (<a href="" class="xref" data-knowl="./knowl/eq2_19.html" title="Equation 2.5.3">(2.5.3)</a>), one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_19.html">
\begin{equation*}
x=-\frac{m}{k} \left(v_0+\frac{mg}{k}\right) e^{-\frac{k}{m}t}-\frac{mg}{k}t+D.
\end{equation*}
</div>
<p class="continuation">The initial condition gives</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_19.html">
\begin{equation*}
D=\frac{m}{k} \left(v_0+\frac{mg}{k}\right).
\end{equation*}
</div>
<p class="continuation">Thus, the solution to the displacement is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_19.html" id="eq2_20">
\begin{equation}
x=\frac{m}{k} \left( v_0+\frac{mg}{k} \right) \left( 1-e^{-\frac{k}{m}t}  \right)-\frac{mg}{k}t.\tag{2.5.4}
\end{equation}
</div>
<p id="p-38">When the velocity is zero, the body reaches the maximum height. In (<a href="" class="xref" data-knowl="./knowl/eq2_18_1.html" title="Equation 2.5.2">(2.5.2)</a>), setting <span class="process-math">\(v=0\text{,}\)</span> one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_18_1.html ./knowl/eq2_20.html">
\begin{equation*}
0=\left(v_0+\frac{mg}{k}\right) e^{-\frac{k}{m}t}-\frac{mg}{k},
\end{equation*}
</div>
<p class="continuation">which leads to</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_18_1.html ./knowl/eq2_20.html">
\begin{equation*}
t=\frac{m}{k} \ln \left(\frac{v_0 k}{mg}+1\right).
\end{equation*}
</div>
<p class="continuation">This is the time when the body reaches the maximum height. And the maximum height is obtained by taking this time into (<a href="" class="xref" data-knowl="./knowl/eq2_20.html" title="Equation 2.5.4">(2.5.4)</a>), one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_18_1.html ./knowl/eq2_20.html">
\begin{equation*}
x=\frac{m v_0}{k}-\frac{m^2 g}{k^2} \ln \left(\frac{v_0 k}{mg}+1 \right).
\end{equation*}
</div>
<p id="p-39"><dfn class="terminology">Example 2</dfn> A body of constant mass <span class="process-math">\(m\)</span> is projected vertically upward from the surface of the earth with an initial velocity <span class="process-math">\(v_0\text{.}\)</span> Neglecting the air resistance but taking into account the variation of the earth gravitational field with altitude. Finda) The expression for the velocity during the motion;b) The maximum height attained by the body;c) The smallest initial velocity for which the body will not return to earth.<dfn class="terminology">Solution:</dfn> From Newton’s second law, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_22.html ./knowl/eq2_21.html ./knowl/eq2_23.html" id="eq2_21">
\begin{equation}
\mathrm{Gravity}=-\frac{mgR^2}{(R+x)^2}=m\frac{\textrm{d} v}{\textrm{d} t}.\tag{2.5.5}
\end{equation}
</div>
<p class="continuation">There are two unknown functions <span class="process-math">\(x(t)\)</span> and <span class="process-math">\(v(t)\)</span> in the above differential equation. And there exists a relation between these two unknown functions</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_22.html ./knowl/eq2_21.html ./knowl/eq2_23.html" id="eq2_22">
\begin{equation}
\frac{\textrm{d} v}{\textrm{d} t}=\frac{\textrm{d} x}{\textrm{d} t} \frac{\textrm{d} v}{\textrm{d} x}=v \frac{\textrm{d} v}{\textrm{d} x}.\tag{2.5.6}
\end{equation}
</div>
<p class="continuation">Taking (<a href="" class="xref" data-knowl="./knowl/eq2_22.html" title="Equation 2.5.6">(2.5.6)</a>) into (<a href="" class="xref" data-knowl="./knowl/eq2_21.html" title="Equation 2.5.5">(2.5.5)</a>), one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_22.html ./knowl/eq2_21.html ./knowl/eq2_23.html" id="eq2_23">
\begin{equation}
-\frac{gR^2}{(R+x)^2}=v\frac{\textrm{d} v}{\textrm{d} x}~\rightarrow~-\frac{gR^2 \textrm{d} x}{(R+x)^2}=v \textrm{d} v~\rightarrow~\frac{gR^2}{R+x}+C=\frac{v^2}{2}.\tag{2.5.7}
\end{equation}
</div>
<p class="continuation">Initially, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_22.html ./knowl/eq2_21.html ./knowl/eq2_23.html">
\begin{equation*}
x(0)=0, \quad v(0)=v_0.
\end{equation*}
</div>
<p class="continuation">This implies at <span class="process-math">\(x=0\text{,}\)</span> <span class="process-math">\(v=v_0\)</span> which further gives</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_22.html ./knowl/eq2_21.html ./knowl/eq2_23.html">
\begin{equation*}
\frac{gR^2}{R+0}+C=\frac{1}{2}v_0^2~\rightarrow~C=\frac{1}{2} v_0^2-gR.
\end{equation*}
</div>
<p class="continuation">From (<a href="" class="xref" data-knowl="./knowl/eq2_23.html" title="Equation 2.5.7">(2.5.7)</a>), one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_22.html ./knowl/eq2_21.html ./knowl/eq2_23.html">
\begin{equation*}
\frac{1}{2} v^2=\frac{gR^2}{R+x}+\frac{1}{2} v_0^2-gR~\rightarrow~v=\pm \sqrt{\frac{2gR^2}{R+x}+v_0^2-2gR}.
\end{equation*}
</div>
<p id="p-40">At the maximum height, denoted by <span class="process-math">\(\xi\text{,}\)</span> the velocity is zero:</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_24.html" id="eq2_24">
\begin{equation}
0=\pm  \sqrt{\frac{2gR^2}{R+\xi}+v_0^2-2gR}~\rightarrow~\xi=\frac{v_0^2R}{2gR-v_0^2}.\tag{2.5.8}
\end{equation}
</div>
<p class="continuation">Not returning to the earth implies <span class="process-math">\(\xi~\rightarrow~\infty\text{.}\)</span> In (<a href="" class="xref" data-knowl="./knowl/eq2_24.html" title="Equation 2.5.8">(2.5.8)</a>), letting <span class="process-math">\(\xi~\rightarrow~\infty\text{,}\)</span> we must have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_24.html">
\begin{equation*}
2gR-v_0^2=0~\rightarrow~v_0=\sqrt{2gR},
\end{equation*}
</div>
<p class="continuation">which is the smallest initial velocity. For <span class="process-math">\(g=9.8m/s^2, R\approx 6286224.5m\text{,}\)</span> <span class="process-math">\(v_0=11100 m/s\)</span> or <span class="process-math">\(v_0=11.1 km/s\text{.}\)</span></p></section></div></main>
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